BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

If a number N is divisible by 33, what will be the minimum

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 416
Joined: Thu Oct 15, 2009 11:52 am
Thanked: 27 times

by regor60 » Thu Feb 01, 2018 6:26 am
VJesus12 wrote:If a number N is divisible by 33, what will be the minimum value of k, such that N^k should definitely be divisible by 27?

A. 1
B. 2
C. 3
D. 4
E. 6

The OA is the option C.

Experts, how can I show that C is the correct answer? May you help me here? I'd be thankful.
For N to be divisible by 33 means N/(3*11) = some integer, call it X.

So N = 3*11*X. Raising N to the k means:

N^k = 3^k*11^k*X^k. Dividing this by 27 means dividing by 3^3, so

N^k/(3^3) = (3^k*11^k*X^k)/3^3

Now, there could be other powers of 3 hiding in the X depending on what numbers are actually used, but to guarantee that 3^3 can divide into the numerator, 3^k has to be at least as great as 3^3, meaning that k = 3, C
Top
Senior | Next Rank: 100 Posts
Posts: 82
Joined: Mon Jan 15, 2018 2:01 am

by DrMaths » Thu Feb 01, 2018 9:00 am
N is divisible by 33, so N = 33*f (where f is an unknown factor) = 3*11*f
N^k is divisible by 27, so N^k = 27*g (where g is an unknown factor) = 3*3*3*g = (3^3)*g
Combining these 2 facts, we get
[3*11*f]^k = (3^3)*g
3^k * (11*f)^k = (3^3)*g
Comparing similar bases (namely powers of 3 in this case), we see that 3^k = 3^3 (minimum)
Comparing powers we see that k = 3 also
ANSWER = C
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8086
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Fri Feb 02, 2018 11:27 am
VJesus12 wrote:If a number N is divisible by 33, what will be the minimum value of k, such that N^k should definitely be divisible by 27?

A. 1
B. 2
C. 3
D. 4
E. 6
Since N is divisible by 33, we see that N has, at a minimum, one prime factor of 3. In order for N^k/27 = integer, we see that N must have at least three primes of 3 (since 27 has three factors of 3), and thus the minimum value of k would have to be 3, since 33^3 = 3^3 x 11^3 = 27 x 11^3.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Post new topic Post Reply
• Page 1 of 1